Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $q = \dfrac{-4k - 16}{k^2 + 5k + 4} \div \dfrac{9k - 81}{-9k - 9} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-4k - 16}{k^2 + 5k + 4} \times \dfrac{-9k - 9}{9k - 81} $ First factor the quadratic. $q = \dfrac{-4k - 16}{(k + 1)(k + 4)} \times \dfrac{-9k - 9}{9k - 81} $ Then factor out any other terms. $q = \dfrac{-4(k + 4)}{(k + 1)(k + 4)} \times \dfrac{-9(k + 1)}{9(k - 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ -4(k + 4) \times -9(k + 1) } { (k + 1)(k + 4) \times 9(k - 9) } $ $q = \dfrac{ 36(k + 4)(k + 1)}{ 9(k + 1)(k + 4)(k - 9)} $ Notice that $(k + 4)$ and $(k + 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ 36(k + 4)\cancel{(k + 1)}}{ 9\cancel{(k + 1)}(k + 4)(k - 9)} $ We are dividing by $k + 1$ , so $k + 1 \neq 0$ Therefore, $k \neq -1$ $q = \dfrac{ 36\cancel{(k + 4)}\cancel{(k + 1)}}{ 9\cancel{(k + 1)}\cancel{(k + 4)}(k - 9)} $ We are dividing by $k + 4$ , so $k + 4 \neq 0$ Therefore, $k \neq -4$ $q = \dfrac{36}{9(k - 9)} $ $q = \dfrac{4}{k - 9} ; \space k \neq -1 ; \space k \neq -4 $